2007 AMC 12A Problem 12

Below is the professionally curated solution for Problem 12 of the 2007 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

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Concepts:parity basic probability

Difficulty rating: 1410

12.

Integers $a,$ $b,$ $c,$ and $d,$ not necessarily distinct, are chosen independently and at random from $0$ to $2007,$ inclusive. What is the probability that $ad-bc$ is even?

$\dfrac{3}{8}$

$\dfrac{7}{16}$

$\dfrac{1}{2}$

$\dfrac{9}{16}$

$\dfrac{5}{8}$

Solution:

Exactly half of the integers from $0$ to $2007$ are odd.

A product $ad$ is odd only when both factors are odd, with probability $\tfrac12\cdot\tfrac12=\tfrac14,$ and even with probability $\tfrac34.$ The same holds for $bc.$

Then $ad-bc$ is even when both products are odd or both are even: $\tfrac14\cdot\tfrac14+\tfrac34\cdot\tfrac34=\tfrac{10}{16}=\tfrac58.$

Thus, the correct answer is E.

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